题目
一个质点同时参与两个在同一直线上的简谐振动,其表达式分别为_(1)=4times (10)^-2cos (2t+dfrac (1)(6)pi ), _(1)=4times (10)^-2cos (2t+dfrac (1)(6)pi ) (SI)则其合成振动的振幅为____ _______ ___,初相为___________________.
一个质点同时参与两个在同一直线上的简谐振动,其表达式分别为
, 
(SI)
则其合成振动的振幅为____ _______ ___,初相为___________________.
题目解答
答案
1×10-2 m ; /6
解析
步骤 1：确定两个简谐振动的振幅和相位
给定的两个简谐振动表达式分别为：
${x}_{1}=4\times {10}^{-2}\cos (2t+\dfrac {1}{6}\pi )$
${x}_{2}=3\times {10}^{-2}\cos (2t-\dfrac {5}{6}\pi )$
其中，${x}_{1}$的振幅为$4\times {10}^{-2}$，相位为$\dfrac {1}{6}\pi$；${x}_{2}$的振幅为$3\times {10}^{-2}$，相位为$-\dfrac {5}{6}\pi$。
步骤 2：计算合成振动的振幅
合成振动的振幅$A$可以通过以下公式计算：
$A=\sqrt{{A}_{1}^{2}+{A}_{2}^{2}+2{A}_{1}{A}_{2}\cos(\phi_{1}-\phi_{2})}$
其中，${A}_{1}=4\times {10}^{-2}$，${A}_{2}=3\times {10}^{-2}$，$\phi_{1}=\dfrac {1}{6}\pi$，$\phi_{2}=-\dfrac {5}{6}\pi$。
代入公式得：
$A=\sqrt{(4\times {10}^{-2})^{2}+(3\times {10}^{-2})^{2}+2\times (4\times {10}^{-2})\times (3\times {10}^{-2})\times \cos(\dfrac {1}{6}\pi-(-\dfrac {5}{6}\pi))}$
$A=\sqrt{(4\times {10}^{-2})^{2}+(3\times {10}^{-2})^{2}+2\times (4\times {10}^{-2})\times (3\times {10}^{-2})\times \cos(\pi)}$
$A=\sqrt{(4\times {10}^{-2})^{2}+(3\times {10}^{-2})^{2}-2\times (4\times {10}^{-2})\times (3\times {10}^{-2})}$
$A=\sqrt{(4\times {10}^{-2})^{2}+(3\times {10}^{-2})^{2}-2\times (4\times {10}^{-2})\times (3\times {10}^{-2})}$
$A=\sqrt{16\times {10}^{-4}+9\times {10}^{-4}-24\times {10}^{-4}}$
$A=\sqrt{1\times {10}^{-4}}$
$A=1\times {10}^{-2}$
步骤 3：计算合成振动的初相
合成振动的初相$\phi$可以通过以下公式计算：
$\tan\phi=\dfrac{{A}_{1}\sin\phi_{1}+{A}_{2}\sin\phi_{2}}{{A}_{1}\cos\phi_{1}+{A}_{2}\cos\phi_{2}}$
代入公式得：
$\tan\phi=\dfrac{(4\times {10}^{-2})\sin(\dfrac {1}{6}\pi)+(3\times {10}^{-2})\sin(-\dfrac {5}{6}\pi)}{(4\times {10}^{-2})\cos(\dfrac {1}{6}\pi)+(3\times {10}^{-2})\cos(-\dfrac {5}{6}\pi)}$
$\tan\phi=\dfrac{(4\times {10}^{-2})\times \dfrac {1}{2}+(3\times {10}^{-2})\times (-\dfrac {1}{2})}{(4\times {10}^{-2})\times \dfrac {\sqrt{3}}{2}+(3\times {10}^{-2})\times (-\dfrac {\sqrt{3}}{2})}$
$\tan\phi=\dfrac{2\times {10}^{-2}-1.5\times {10}^{-2}}{2\sqrt{3}\times {10}^{-2}-1.5\sqrt{3}\times {10}^{-2}}$
$\tan\phi=\dfrac{0.5\times {10}^{-2}}{0.5\sqrt{3}\times {10}^{-2}}$
$\tan\phi=\dfrac{1}{\sqrt{3}}$
$\phi=\dfrac {1}{6}\pi$
给定的两个简谐振动表达式分别为：
${x}_{1}=4\times {10}^{-2}\cos (2t+\dfrac {1}{6}\pi )$
${x}_{2}=3\times {10}^{-2}\cos (2t-\dfrac {5}{6}\pi )$
其中，${x}_{1}$的振幅为$4\times {10}^{-2}$，相位为$\dfrac {1}{6}\pi$；${x}_{2}$的振幅为$3\times {10}^{-2}$，相位为$-\dfrac {5}{6}\pi$。
步骤 2：计算合成振动的振幅
合成振动的振幅$A$可以通过以下公式计算：
$A=\sqrt{{A}_{1}^{2}+{A}_{2}^{2}+2{A}_{1}{A}_{2}\cos(\phi_{1}-\phi_{2})}$
其中，${A}_{1}=4\times {10}^{-2}$，${A}_{2}=3\times {10}^{-2}$，$\phi_{1}=\dfrac {1}{6}\pi$，$\phi_{2}=-\dfrac {5}{6}\pi$。
代入公式得：
$A=\sqrt{(4\times {10}^{-2})^{2}+(3\times {10}^{-2})^{2}+2\times (4\times {10}^{-2})\times (3\times {10}^{-2})\times \cos(\dfrac {1}{6}\pi-(-\dfrac {5}{6}\pi))}$
$A=\sqrt{(4\times {10}^{-2})^{2}+(3\times {10}^{-2})^{2}+2\times (4\times {10}^{-2})\times (3\times {10}^{-2})\times \cos(\pi)}$
$A=\sqrt{(4\times {10}^{-2})^{2}+(3\times {10}^{-2})^{2}-2\times (4\times {10}^{-2})\times (3\times {10}^{-2})}$
$A=\sqrt{(4\times {10}^{-2})^{2}+(3\times {10}^{-2})^{2}-2\times (4\times {10}^{-2})\times (3\times {10}^{-2})}$
$A=\sqrt{16\times {10}^{-4}+9\times {10}^{-4}-24\times {10}^{-4}}$
$A=\sqrt{1\times {10}^{-4}}$
$A=1\times {10}^{-2}$
步骤 3：计算合成振动的初相
合成振动的初相$\phi$可以通过以下公式计算：
$\tan\phi=\dfrac{{A}_{1}\sin\phi_{1}+{A}_{2}\sin\phi_{2}}{{A}_{1}\cos\phi_{1}+{A}_{2}\cos\phi_{2}}$
代入公式得：
$\tan\phi=\dfrac{(4\times {10}^{-2})\sin(\dfrac {1}{6}\pi)+(3\times {10}^{-2})\sin(-\dfrac {5}{6}\pi)}{(4\times {10}^{-2})\cos(\dfrac {1}{6}\pi)+(3\times {10}^{-2})\cos(-\dfrac {5}{6}\pi)}$
$\tan\phi=\dfrac{(4\times {10}^{-2})\times \dfrac {1}{2}+(3\times {10}^{-2})\times (-\dfrac {1}{2})}{(4\times {10}^{-2})\times \dfrac {\sqrt{3}}{2}+(3\times {10}^{-2})\times (-\dfrac {\sqrt{3}}{2})}$
$\tan\phi=\dfrac{2\times {10}^{-2}-1.5\times {10}^{-2}}{2\sqrt{3}\times {10}^{-2}-1.5\sqrt{3}\times {10}^{-2}}$
$\tan\phi=\dfrac{0.5\times {10}^{-2}}{0.5\sqrt{3}\times {10}^{-2}}$
$\tan\phi=\dfrac{1}{\sqrt{3}}$
$\phi=\dfrac {1}{6}\pi$